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A body moving with uniform retardation covers 3km before it`s speed is reduced to half of its initial value it comes to rest in another distance of ?
A body moving with uniform retardation covers 3km before it`s speed is reduced to half of its initial value it comes to  rest in another distance of ?

```
4 years ago

Vikas TU
14149 Points
```							using 3rd law of eqn.(u/2)^2 = u^2 – 2*a*3000-3u^2/4 = – 6000au^2 = 8000a...................(1)when it comes to rest,0^2 = (u/2)^2 – 2as2as = u^2/4as = u^2/8ors = u^2/8a..............(2)put eqn. (1) in eqn. (2)and get s value.
```
4 years ago
Ankit Jaiswal
165 Points
```							let us suppose that its initial velocity = uusing equation of motionv2=u2+2as     (u/2)2 = u2 =2*a*3000 (value of final velocity is haif he initial taht is u/2 and s = 3 km =3000m)$\therefore$ (u2/4) – u2 =2*a*3000$\therefore$ (-3u2/4)=2*a*3000$\therefore$ – u2=(2*a*3000*4)/3$\therefore$ u2 = – 8000*a                      ….............................1by appling the same equation when it comes to rest we get     02=(u/2)2 +2*a*s$\therefore$ – u2/4 =2*a*s$\therefore$ – u2/(4*2*a) = s$\therefore$ – u2/8a = safter substituting values form 1 we get$\therefore$ – (-8000*a)/8a = s$\therefore$ s = 1000m = 1kmi think this might help  plz approve if u found it useful thanks
```
4 years ago
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